Problem: $\overline{AC}$ is $8$ units long $\overline{BC}$ is $6$ units long $\overline{AB}$ is $10$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $8$ $6$ $10$
Solution: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 6$ hypotenuse $= \overline{AB} = 10$ $\sin(\angle BAC)=\frac{6}{10}$ $=\dfrac{3}{5}$